A body thrown vertically upwards from the gro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The maximum height reached by a body thrown with initial velocity $u$ is given by $H = \frac{u^2}{2g}$,which implies $u^2 = 2gH$.Using the equatio

Vedclass · Accepted Answer

$3: 2$

Vedclass · Answer

(C) The maximum height reached by a body thrown with initial velocity $u$ is given by $H = \frac{u^2}{2g}$,which implies $u^2 = 2gH$.Using the equation of motion $v^2 = u^2 - 2gh$,we calculate the velocity at any height $h$.For height $h_1 = \frac{3H}{4}$,the velocity $v_1$ is given by $v_1^2 = u^2 - 2g(\frac{3H}{4}) = 2gH - \frac{3gH}{2} = \frac{gH}{2}$.For height $h_2 = \frac{8H}{9}$,the velocity $v_2$ is given by $v_2^2 = u^2 - 2g(\frac{8H}{9}) = 2gH - \frac{16gH}{9} = \frac{2gH}{9}$.The ratio of the velocities is $\frac{v_1}{v_2} = \sqrt{\frac{v_1^2}{v_2^2}} = \sqrt{\frac{gH/2}{2gH/9}} = \sqrt{\frac{1}{2} 	imes \frac{9}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

$A$ body thrown vertically upwards from the ground reaches a maximum height $H$. The ratio of the velocities of the body at heights $\frac{3H}{4}$ and $\frac{8H}{9}$ from the ground is

Similar Questions

$A$ ball is projected upwards. Its acceleration at the highest point is

$A$ ball is dropped on the floor from a height of $10 \, m$. It rebounds to a height of $2.5 \, m$. If the ball is in contact with the floor for $0.01 \, s$,the average acceleration during contact is:

What is the speed-time graph for a ball thrown vertically upwards? (Neglect air resistance)

$A$ body dropped from a height $h$ with an initial speed zero,strikes the ground with a velocity $3 \ km/h$. Another body of the same mass is dropped from the same height $h$ with an initial speed of $4 \ km/h$. Find the final velocity of the second body with which it strikes the ground in $km/h$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module